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If [tex]\(2a - b + c = 0\)[/tex], show that [tex]\(4a^2 - b^2 + c^2 + 4ac = 0\)[/tex].

Sagot :

GinnyAnswer
Sure, let's solve this problem step-by-step.

Given the equation:
[tex]\[ 2a - b + c = 0 \][/tex]

We need to show that:
[tex]\[ 4a^2 - b^2 + c^2 + 4ac = 0 \][/tex]

First, we rearrange the given equation to express [tex]\( b \)[/tex] in terms of [tex]\( a \)[/tex] and [tex]\( c \)[/tex].
[tex]\[ 2a - b + c = 0 \][/tex]

Rearranging the terms, we get:
[tex]\[ b = 2a + c \][/tex]

Now, substitute [tex]\( b = 2a + c \)[/tex] into the expression [tex]\( 4a^2 - b^2 + c^2 + 4ac \)[/tex]:

[tex]\[ 4a^2 - (2a + c)^2 + c^2 + 4ac \][/tex]

Let's expand the square term [tex]\( (2a + c)^2 \)[/tex]:
[tex]\[ (2a + c)^2 = 4a^2 + 4ac + c^2 \][/tex]

Substitute this back into the expression:
[tex]\[ 4a^2 - (4a^2 + 4ac + c^2) + c^2 + 4ac \][/tex]

Now, distribute the negative sign:
[tex]\[ 4a^2 - 4a^2 - 4ac - c^2 + c^2 + 4ac \][/tex]

Combine like terms:
[tex]\[ 4a^2 - 4a^2 - 4ac - c^2 + c^2 + 4ac = 0 \][/tex]

Notice that [tex]\( 4a^2 - 4a^2 \)[/tex], [tex]\(-4ac + 4ac\)[/tex], and [tex]\(-c^2 + c^2\)[/tex] all cancel out, leaving us with:
[tex]\[ 0 = 0 \][/tex]

Therefore, we have shown that:
[tex]\[ 4a^2 - b^2 + c^2 + 4ac = 0 \][/tex]

as required.
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